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Question

# Let A and B be any two points on the lines represented by 4x2−9y2=0. If the area of triangle OAB is 5 (O is origin) then which of the following is the possible equation of the locus of midpoint of AB ?

A
4x2+9y2+30=0
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B
4x29y230=0
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C
9x24y230=0
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D
9x2+4y230=0
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Solution

## The correct option is B 4x2−9y2−30=0 Area of △OAB=∣∣∣12OA.OBsinθ ∣∣∣ =5 OA=√133h ; OB=√133k 5=∣∣∣12.√133h.√133k.sinθ∣∣∣ tanθ=23−(−23)1+23−(−23)=125⇒sinθ=1213 ⇒|hk|=152⇒hk=±152 Midpoint of AB=⎛⎜ ⎜ ⎜⎝h+k2,2h3−2k32⎞⎟ ⎟ ⎟⎠ =(h+k2,h−k3) ⇒h+k=2x; h−k=3y ⇒2h=2x+3y ⇒2k=2x−3y ⇒4hk=4x2−9y2 ⇒4x2−9y2=±30 Alternate Solution: Area of △OAB=12∣∣ ∣ ∣ ∣ ∣∣10 01k−2k31h 2h3∣∣ ∣ ∣ ∣ ∣∣ ⇒5=12∣∣∣2kh3+2kh3∣∣∣ ⇒|hk|=152⇒hk=±152 Midpoint of AB=⎛⎜ ⎜ ⎜⎝h+k2,2h3−2k32⎞⎟ ⎟ ⎟⎠ =(h+k2,h−k3) ⇒h+k=2x; h−k=3y ⇒2h=2x+3y ⇒2k=2x−3y ⇒4hk=4x2−9y2 ⇒4x2−9y2=±30

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